Answer a, b, & c with show of work

Question

A projectile is shot upward from 10 meters above the surface of earth with an initial velocity of 50 meters per second. It's position s(t) measured in meters at time t seconds later is given by s(t) = -4.9 t^2 + 50 t + 10.

(a) write a formula for the average velocity of the ball from 1 second to t seconds.

(b) find the average velocity for the time interval beginning at t = 1 and ending at t = 2.5 seconds; 1.5 seconds; and 1.05 seconds.

(c) what is your estimate for the instantaneous velocity of the ball at t = 1?

Answer 1

A) v=d/t

In this case d=(-4.9t^2+50t+10)/t

B) v=(-4.9t^2+50t+10)/t=

t(50)= " " " =1.49=t

t(1)=55.1

t(2.5)=41.75

C) Depends on the force.Velocity-up=Vy=Voy-gt

g=32.2ft/sec^2; or 9.81m/sec^2

Answer 2

(a) s(t)= -4.9 t^2 + 50 t + 10

Average velocity is (distance traveled)/time = (s(t)-s(1))/(t-1)=-4.9 t^2 + 50 t + 10-(-4.9+50+10)

=-4.9 t^2 + 50t+10+4.9-50-10

=-4.9 t^2 + 50t+-45.1

(b)

substitute the times given in the formula for (a)

(c)

Instantaneous velocity =ds/dt=-9.8t+50

for t=1 this is 40.2m/s